Problem: In the diagram below, $WXYZ$ is a trapezoid such that $\overline{WX}\parallel \overline{ZY}$ and $\overline{WY}\perp\overline{ZY}$.  If $YZ = 12$, $\tan Z = 1.5$, and $\tan X = 2$, then what is $XY$?

[asy]

pair WW,X,Y,Z;

Z = (0,0);

Y = (12,0);

WW = (12,18);

X= (18,18);

draw(WW--X--Y--Z--WW);

label("$W$",WW,N);

label("$X$",X,N);

label("$Y$",Y,S);

label("$Z$",Z,S);

[/asy]
Answer: [asy]
pair WW,X,Y,Z;
Z = (0,0);
Y = (12,0);
WW = (12,18);
X= (18,18);
draw(WW--Y);
draw(rightanglemark(WW,Y,Z,30));
draw(rightanglemark(Y,WW,X,30));
draw(WW--X--Y--Z--WW);
label("$W$",WW,N);
label("$X$",X,N);
label("$Y$",Y,S);
label("$Z$",Z,S);
label("$12$",Y/2,S);
[/asy]

We add $\overline{WY}$ to our diagram and note that because $\overline{WX}\parallel\overline{ZY}$ and $\overline{WY}\perp\overline{ZY}$, we have $\overline{WY}\perp\overline{WX}$.  Therefore, triangles $WYX$ and $WYZ$ are right triangles.

From right triangle $WYZ$, we have $\tan Z = \frac{WY}{ZY}$, so $WY = ZY\cdot \tan Z = 12\cdot 1.5 = 18$.  From right triangle $WXY$, we have $\tan X = \frac{WY}{WX}$, so \[WX = \frac{WY}{\tan X} = \frac{18}{2} =9.\]Finally, the Pythagorean Theorem gives  \begin{align*}
XY&=\sqrt{WY^2 + WX^2} \\
&= \sqrt{18^2 + 9^2} \\
&= \sqrt{(2\cdot 9)^2 + 9^2} \\
&= \sqrt{5\cdot 9^2} \\
&= \boxed{9\sqrt{5}}.
\end{align*}